Obliczmy granicę:

$\underset{n\to \infty }{\lim }\left(\frac{11n^3+6n+5}{6n^3+1}-\frac{2n^2+2n+1}{5n^2-4}\right)=\underset{n\to \infty }{\lim }\left(\frac{n^3\left(11+\frac{6}{n^2}+\frac{5}{n^3}\right)}{n^3\left(6+\frac{1}{n^3}\right)}-\frac{n^2\left(2+\frac{2}{n}+\frac{1}{n^2}\right)}{n^2\left(5-\frac{4}{n^2}\right)}\right)=\underset{n\to \infty }{\lim }\left(\frac{11+\frac{6}{n^2}+\frac{5}{n^3}}{6+\frac{1}{n^3}}-\frac{2+\frac{2}{n}+\frac{1}{n^2}}{5-\frac{4}{n^2}}\right)=\frac{11}{6}-\frac{2}{5}=\frac{55}{30}-\frac{12}{30}=\frac{43}{30}=1,43\dots$  

Kodujemy

$143$